-0.000 000 000 742 147 676 646 708 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 708 93(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 676 646 708 93(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 676 646 708 93| = 0.000 000 000 742 147 676 646 708 93


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 147 676 646 708 93.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 742 147 676 646 708 93 × 2 = 0 + 0.000 000 001 484 295 353 293 417 86;
  • 2) 0.000 000 001 484 295 353 293 417 86 × 2 = 0 + 0.000 000 002 968 590 706 586 835 72;
  • 3) 0.000 000 002 968 590 706 586 835 72 × 2 = 0 + 0.000 000 005 937 181 413 173 671 44;
  • 4) 0.000 000 005 937 181 413 173 671 44 × 2 = 0 + 0.000 000 011 874 362 826 347 342 88;
  • 5) 0.000 000 011 874 362 826 347 342 88 × 2 = 0 + 0.000 000 023 748 725 652 694 685 76;
  • 6) 0.000 000 023 748 725 652 694 685 76 × 2 = 0 + 0.000 000 047 497 451 305 389 371 52;
  • 7) 0.000 000 047 497 451 305 389 371 52 × 2 = 0 + 0.000 000 094 994 902 610 778 743 04;
  • 8) 0.000 000 094 994 902 610 778 743 04 × 2 = 0 + 0.000 000 189 989 805 221 557 486 08;
  • 9) 0.000 000 189 989 805 221 557 486 08 × 2 = 0 + 0.000 000 379 979 610 443 114 972 16;
  • 10) 0.000 000 379 979 610 443 114 972 16 × 2 = 0 + 0.000 000 759 959 220 886 229 944 32;
  • 11) 0.000 000 759 959 220 886 229 944 32 × 2 = 0 + 0.000 001 519 918 441 772 459 888 64;
  • 12) 0.000 001 519 918 441 772 459 888 64 × 2 = 0 + 0.000 003 039 836 883 544 919 777 28;
  • 13) 0.000 003 039 836 883 544 919 777 28 × 2 = 0 + 0.000 006 079 673 767 089 839 554 56;
  • 14) 0.000 006 079 673 767 089 839 554 56 × 2 = 0 + 0.000 012 159 347 534 179 679 109 12;
  • 15) 0.000 012 159 347 534 179 679 109 12 × 2 = 0 + 0.000 024 318 695 068 359 358 218 24;
  • 16) 0.000 024 318 695 068 359 358 218 24 × 2 = 0 + 0.000 048 637 390 136 718 716 436 48;
  • 17) 0.000 048 637 390 136 718 716 436 48 × 2 = 0 + 0.000 097 274 780 273 437 432 872 96;
  • 18) 0.000 097 274 780 273 437 432 872 96 × 2 = 0 + 0.000 194 549 560 546 874 865 745 92;
  • 19) 0.000 194 549 560 546 874 865 745 92 × 2 = 0 + 0.000 389 099 121 093 749 731 491 84;
  • 20) 0.000 389 099 121 093 749 731 491 84 × 2 = 0 + 0.000 778 198 242 187 499 462 983 68;
  • 21) 0.000 778 198 242 187 499 462 983 68 × 2 = 0 + 0.001 556 396 484 374 998 925 967 36;
  • 22) 0.001 556 396 484 374 998 925 967 36 × 2 = 0 + 0.003 112 792 968 749 997 851 934 72;
  • 23) 0.003 112 792 968 749 997 851 934 72 × 2 = 0 + 0.006 225 585 937 499 995 703 869 44;
  • 24) 0.006 225 585 937 499 995 703 869 44 × 2 = 0 + 0.012 451 171 874 999 991 407 738 88;
  • 25) 0.012 451 171 874 999 991 407 738 88 × 2 = 0 + 0.024 902 343 749 999 982 815 477 76;
  • 26) 0.024 902 343 749 999 982 815 477 76 × 2 = 0 + 0.049 804 687 499 999 965 630 955 52;
  • 27) 0.049 804 687 499 999 965 630 955 52 × 2 = 0 + 0.099 609 374 999 999 931 261 911 04;
  • 28) 0.099 609 374 999 999 931 261 911 04 × 2 = 0 + 0.199 218 749 999 999 862 523 822 08;
  • 29) 0.199 218 749 999 999 862 523 822 08 × 2 = 0 + 0.398 437 499 999 999 725 047 644 16;
  • 30) 0.398 437 499 999 999 725 047 644 16 × 2 = 0 + 0.796 874 999 999 999 450 095 288 32;
  • 31) 0.796 874 999 999 999 450 095 288 32 × 2 = 1 + 0.593 749 999 999 998 900 190 576 64;
  • 32) 0.593 749 999 999 998 900 190 576 64 × 2 = 1 + 0.187 499 999 999 997 800 381 153 28;
  • 33) 0.187 499 999 999 997 800 381 153 28 × 2 = 0 + 0.374 999 999 999 995 600 762 306 56;
  • 34) 0.374 999 999 999 995 600 762 306 56 × 2 = 0 + 0.749 999 999 999 991 201 524 613 12;
  • 35) 0.749 999 999 999 991 201 524 613 12 × 2 = 1 + 0.499 999 999 999 982 403 049 226 24;
  • 36) 0.499 999 999 999 982 403 049 226 24 × 2 = 0 + 0.999 999 999 999 964 806 098 452 48;
  • 37) 0.999 999 999 999 964 806 098 452 48 × 2 = 1 + 0.999 999 999 999 929 612 196 904 96;
  • 38) 0.999 999 999 999 929 612 196 904 96 × 2 = 1 + 0.999 999 999 999 859 224 393 809 92;
  • 39) 0.999 999 999 999 859 224 393 809 92 × 2 = 1 + 0.999 999 999 999 718 448 787 619 84;
  • 40) 0.999 999 999 999 718 448 787 619 84 × 2 = 1 + 0.999 999 999 999 436 897 575 239 68;
  • 41) 0.999 999 999 999 436 897 575 239 68 × 2 = 1 + 0.999 999 999 998 873 795 150 479 36;
  • 42) 0.999 999 999 998 873 795 150 479 36 × 2 = 1 + 0.999 999 999 997 747 590 300 958 72;
  • 43) 0.999 999 999 997 747 590 300 958 72 × 2 = 1 + 0.999 999 999 995 495 180 601 917 44;
  • 44) 0.999 999 999 995 495 180 601 917 44 × 2 = 1 + 0.999 999 999 990 990 361 203 834 88;
  • 45) 0.999 999 999 990 990 361 203 834 88 × 2 = 1 + 0.999 999 999 981 980 722 407 669 76;
  • 46) 0.999 999 999 981 980 722 407 669 76 × 2 = 1 + 0.999 999 999 963 961 444 815 339 52;
  • 47) 0.999 999 999 963 961 444 815 339 52 × 2 = 1 + 0.999 999 999 927 922 889 630 679 04;
  • 48) 0.999 999 999 927 922 889 630 679 04 × 2 = 1 + 0.999 999 999 855 845 779 261 358 08;
  • 49) 0.999 999 999 855 845 779 261 358 08 × 2 = 1 + 0.999 999 999 711 691 558 522 716 16;
  • 50) 0.999 999 999 711 691 558 522 716 16 × 2 = 1 + 0.999 999 999 423 383 117 045 432 32;
  • 51) 0.999 999 999 423 383 117 045 432 32 × 2 = 1 + 0.999 999 998 846 766 234 090 864 64;
  • 52) 0.999 999 998 846 766 234 090 864 64 × 2 = 1 + 0.999 999 997 693 532 468 181 729 28;
  • 53) 0.999 999 997 693 532 468 181 729 28 × 2 = 1 + 0.999 999 995 387 064 936 363 458 56;
  • 54) 0.999 999 995 387 064 936 363 458 56 × 2 = 1 + 0.999 999 990 774 129 872 726 917 12;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 742 147 676 646 708 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 708 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 708 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 646 708 93 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111